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Komok [63]
2 years ago
13

A 7-hp (shaft) pump is used to raise water to an elevation of 15 m. If the mechanical efficiency of the pump is 82 percent, dete

rmine the maximum volume flow rate of water.
Engineering
1 answer:
Natali [406]2 years ago
5 0

The maximum volume flow rate of water is determined as 0.029 m³/s.

<h3>Power of the pump</h3>

The power of the pump is watt is calculated as follows;

1 hp = 745.69 W

7 hp = ?

= 7 x 745.69 W

= 5,219.83 W

<h3>Mass flow rate of water</h3>

η = mgh/P

mgh = ηP

m = ηP/gh

m = (0.82 x 5,219.83)/(9.8 x 15)

m = 29.12 kg/s

<h3>Maximum volume rate</h3>

V = m/ρ

where;

  • ρ is density of water = 1000 kg/m³

V = (29.12)/(1000)

V = 0.029 m³/s

Learn more about volume flow rate here: brainly.com/question/21630019

#SPJ12

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A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr
Hoochie [10]

Answer:

\Delta m = 102.25\,kg

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

m_{o} = \rho_{air}\cdot V_{tank}

m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{o} = 31.25\,kg

The final mass inside the rigid tank is:

m_{f} = \rho \cdot V_{tank}

m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{f}= 133.5\,kg

The supplied air mass is:

\Delta m = m_{f}-m_{o}

\Delta m = 133.5\,kg-31.25\,kg

\Delta m = 102.25\,kg

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3 years ago
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A. 5 seconds :) Good luck!
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3 years ago
Four common causes of product failure are poor design, poor construction, poorly communicated operating instructions, and ______
Inga [223]

Answer:

B. operator error or misuse

Explanation:

A product is a failure if it is not able to achieve the anticipated life cycle as expected by the organization.

In such a case, there is a withdrawal of the product from the market as a result of its ultimate failure of a product to achieve profitability.

Four common causes of product failure are poor design, poor construction, poorly communicated operating instructions, and <u>operator error or misuse</u>

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8 0
3 years ago
: During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter, smooth, concrete storm sewer. If the
Montano1993 [528]

Answer

diameter of parking lot = 18 in

flowrate = 10 ft³/s

pressure drop = 100 ft

using general equation

\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g}+Z_1 = \dfrac{P_2{\gamma} + \dfrac{v_2^2}{2g} + Z_2 +\dfrac{fLV^2}{2\rho D}

V = \dfrac{Q}{A} = \dfrac{10}{\dfrac{\pi}{4}(\dfrac{18}{12})^2} = 5.66\ ft/s

\Delta P = \gamma (Z_2-Z_1) +\dfrac{fLV^2}{2\rho D}

taking f = 0.0185

at Z₁ = Z₂

\Delta P = \dfrac{0.0185 \times 100\times 1.94\times 5.66^2}{2\dfrac{18}{12} (2)}

ΔP = 0.266 psi

b) when flow is uphill z₂-z₁ = 2

\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266

\Delta P= 1.13\ psi

c) When flow is downhill  z₂-z₁ = -2

\Delta P =62.4\times 2 \times \dfrac{1}{144} +0.266

\Delta P=-0.601\ psi

7 0
3 years ago
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